Triangle LOVE
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1033 Accepted Submission(s): 425
Problem Description
Recently, scientists find that there is love between any of two people. For example, between A and B, if A don’t love B, then B must love A, vice versa. And there is no possibility that two people love each other, what a crazy world!
Now, scientists want to know whether or not there is a “Triangle Love” among N people. “Triangle Love” means that among any three people (A,B and C) , A loves B, B loves C and C loves A.
Your problem is writing a program to read the relationship among N people firstly, and return whether or not there is a “Triangle Love”.
Now, scientists want to know whether or not there is a “Triangle Love” among N people. “Triangle Love” means that among any three people (A,B and C) , A loves B, B loves C and C loves A.
Your problem is writing a program to read the relationship among N people firstly, and return whether or not there is a “Triangle Love”.
Input
The first line contains a single integer t (1 <= t <= 15), the number of test cases.
For each case, the first line contains one integer N (0 < N <= 2000).
In the next N lines contain the adjacency matrix A of the relationship (without spaces). Ai,j = 1 means i-th people loves j-th people, otherwise Ai,j = 0.
It is guaranteed that the given relationship is a tournament, that is, Ai,i= 0, Ai,j ≠ Aj,i(1<=i, j<=n,i≠j).
For each case, the first line contains one integer N (0 < N <= 2000).
In the next N lines contain the adjacency matrix A of the relationship (without spaces). Ai,j = 1 means i-th people loves j-th people, otherwise Ai,j = 0.
It is guaranteed that the given relationship is a tournament, that is, Ai,i= 0, Ai,j ≠ Aj,i(1<=i, j<=n,i≠j).
Output
For each case, output the case number as shown and then print “Yes”, if there is a “Triangle Love” among these N people, otherwise print “No”.
Take the sample output for more details.
Take the sample output for more details.
Sample Input
2 5 00100 10000 01001 11101 11000 5 01111 00000 01000 01100 01110
Sample Output
Case #1: Yes Case #2: No
AC代码：
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int t,n;
int flage;
int cnt=1;
char map[2000][2000];
#include<stdlib.h>
#include<string.h>
int t,n;
int flage;
int cnt=1;
char map[2000][2000];
int main()
{
scanf("%d",&t);
while(t--)
{
flage=0;
scanf("%d",&n);
for(int i=0;i<n;i++)
scanf("%s",map[i]);
for(int i=0;i<n;i++)
{
for(int j=0;map[i][j]=='1'&&j<n;j++)//此处不是很明白；
{
for(int k=0;k<n;k++)
{
if(map[i][k]=='0'&&map[j][k]=='1')
{
flage=1;
break;
}
}
if(flage)
break;
}
if(flage)
break;
}
if(flage)
printf("Case #%d: Yes\n",cnt++);
else
printf("Case #%d: No\n",cnt++);
}
system("pause");
return 0;
}
链接：http://acm.hdu.edu.cn/showproblem.php?pid=4324
{
scanf("%d",&t);
while(t--)
{
flage=0;
scanf("%d",&n);
for(int i=0;i<n;i++)
scanf("%s",map[i]);
for(int i=0;i<n;i++)
{
for(int j=0;map[i][j]=='1'&&j<n;j++)//此处不是很明白；
{
for(int k=0;k<n;k++)
{
if(map[i][k]=='0'&&map[j][k]=='1')
{
flage=1;
break;
}
}
if(flage)
break;
}
if(flage)
break;
}
if(flage)
printf("Case #%d: Yes\n",cnt++);
else
printf("Case #%d: No\n",cnt++);
}
system("pause");
return 0;
}
链接：http://acm.hdu.edu.cn/showproblem.php?pid=4324
